Team Quanta gladly presents all possible short questions of BS Physics book Waves and Oscillation Chapter#01: Harmonic Motion.
Q.1 Any real spring has mass. If this mass is taken into account, explain qualitatively how this will affect the period of oscillation of mass spring system.
Answer: Time period of mass spring system is directly proportional to the mass m where m is the mass of the spring is ignored as compared to the mass of the block. But if we consider the mass of the spring also, the total mass increases. This increasing the period of oscillation of the mass spring system. But spring with finite mass will now result in the damping of the motion of the oscillator.
Q.2 How are each of the following properties of a simple harmonic oscillator affected by doubling he amplitude: period spring constant, total mechanical energy, maximum velocity, and maximum acceleration?
Answer:
The period of an oscillator is independent of the amplitude of the oscillator.
Again, since $K=\ \frac{F}{x}$, so doubling the amplitude means doubling the amplitude means doubling the force F. Hence K remains the same.
The total mechanical energy is directly proportional to the square of the amplitude$\ (E\propto\ x_m^2\ )$, so its value increases four times.
Since $v_m=\ {\omega x}_m$, so doubling the amplitude doubles the maximum speed.
As $a_m=\ \omega^2x_m$, so maximum acceleration also becomes doubled with doubling the amplitude.
Q.3 what changes would you make in harmonic oscillator that would double the maximum speed of oscillating mass?
Answer: The maximum speed of an oscillating mass is given by $$v_m=\ {\omega x}_m=\ \sqrt{\frac{K}{m}}x_m$$, hence we can double the maximum speed by;
Either doubling the amplitude or
By increasing four time and
By decreasing mass four times
Q.4 Periodical with arguments whether the period of a pendulum increases or decreases, when its amplitude is increased?
Answer: Increasing the amplitude means that there is a larger distance to travel, but the restoring force also increases, which proportionally increases the acceleration. This means the mass can travel a greater distance at a greater speed. Hence, its time period remains unchanged.
Q.5 why are damping devices often used on machinery? Give an example?
Answer: Damping devices help us in controlling the motion of machinery such as the un-necessary vibrations produced in it shock absorber is an example of damping device.
Q.6 How can a pendulum be used so that it treats out a sinusoidal curve?
Answer: By attaching a pen marker with the bob and making the marker to slide past a paper that is moved against the bob at a steady speed, one can trace the sinusoidal curve due to the motion of the pendulum. Similarly, if the bb of the pendulum is field with sand and the sand is allowed to drain from a narrow hole below the bob we can obtain a sinusoidal trace on a sheet of paper moved below the pendulum.
Q.7 Differentiate between hook’s law and restoring force?
Answer: Hook’s law is the first classical example of an explanation of elasticity, which is the property of an object or material which causes it to be restored to its original shape after distortion. While, restoring force is the to return to a normal shape after experiencing distortion.
Q.8 Predict by qualitative arguments wither the period of pendulum will increase or decrease when its amplitude is increased.
Answer: The period of pendulum is independent of its amplitude of vibration. By increasing the amplitude, its period will remain the same.
Q.9 A spring has a force constant K and a mass m is suspended from it. The spring is cut in half and the same is suspended from one of the Halves. How the frequencies of oscillation, before and after the spring are is cut related.
Answer: The frequency of mass spring oscillation system is;
$$f=\ \frac{1}{2\pi}\sqrt{\frac{K}{m}}$$
From the formula clear that frequency is independent of length of spring.
Q.10 Why does a pendulum hardly swing under water?
Answer: Due to viscosity of water the damping force is large and pendulum hardly swings under water.
Q.11 How can we use the physical pendulum to measure the acceleration due to gravity ‘g’?
Answer: We know that the formula to measure the acceleration due to gravity ’g’ of physical pendulum
$$T=2\pi\ \sqrt{\frac{I}{mgl}}$$
$$T^2=2\pi\ \frac{I}{mgl}$$
$$g=\ \frac{{4\pi}^{2\ \ }I}{{mT}^2\ l\ }$$
Hare ‘I’ is moment of inertia; m is mass and (l) is the length of physical pendulum.
Q.12 What happens to the frequency of a swing as its oscillations die down from large amplitude to small?
Answer: We know that the angular frequency of a damped oscillator which given by;
$$\omega^{\prime\ }=\sqrt{\omega_o^2-\beta^2}$$
$$\omega^\prime=\ \sqrt{\omega_o^2-\left(\frac{b}{2m}\right)^2}$$
$$\omega^\prime=\ \sqrt{\omega_o^2-\ \frac{b^2}{{4m}^2}}$$
Where $\omega_o=\ \sqrt{\frac{K\ }{m}}$ is the frequency with no damping it is obvious from this equation that for small value of damping factor b,$\ \omega$ is nearly equal to $\omega_o$. However, w’ is different from $\omega_o$ for large value of b such that $\omega$ decrease which means that the period of oscillation increases the damping force show down the motion, hence the frequency of vibration decreases as the oscillation die down for large amplitude to small.
Q.13 Define simple harmonic motion?
Answer: The vibrating and periodic motion in which acceleration of the body is directly proportional to the displacement and its direction always directed to-ward the mean position.
$$a\propto-x$$
The total energy of the system during simple harmonic motion remains conserved.
Q.14 When mass m1 is huge from spring A and a smaller mass m2 is huge from B, the springs are stretched by the same distance. If the system is then put into vertical simple harmonic oscillations with the same amplitude, which system will have more energy?
Answer: The total energy of block-spring system is given by;
$$E= \frac{1}{2}{Kx}_m^2$$
Thus with the same amplitude, this energy is directly proportional to the spring constant. The value of the spring constant, $K=\ \frac{F}{x}=\ \frac{mg}{x}$. With the same extension produced $K \propto m$. Since m1>m2 , therefore, $K_A>K_B$. Thus energy stored in system A is more than the energy stored in B.
Q.15 What happens to the motion of an oscillating system if the sign of the force term, -kx were changed?
Answer: The force no longer remains an elastic restoring force. The oscillatory motion will change into translator motion.
Q.16 Could we ever construct a true simple pendulum? Explain your answer.
Answer: A true simple pendulum has the following features:
A heavy point size bob,
A mass less inextensible suspension
A firm frictionless suspension
Neither of these quantities are feasible. Hence a real simple pendulum cannot be realized.
Q.17 Will the frequency of oscillation of a torsion pendulum change if you take it to the moon? What about the frequency of a physical pendulum, such as a wooden plank swinging from the end?
Answer: The frequency of oscillation of a torsional oscillator is $f=\ \frac{1}{2\pi}\sqrt{\frac{K}{I}}$. Where k is the torsional constant of the suspension and I is its moment of inertia. Thus if we take this pendulum to the moon, its period of oscillation will remain same.
$f=\ \frac{1}{2\pi}\sqrt{\frac{mgh}{I}}$ and is function of g. hence if we take it to moon, its frequency of oscillation will decrease.
Q.18 Give some examples of common phenomena in which resonance plays an important role.
Answer: Common examples of phenomena in which resonance plays an important role are,
All musical instruments employ resonance in producing sound
Cooking in microwave oven is still another example in which resonance effect is used.
Q.19 Suppose that a system consists of a block of unknown mass and a spring of unknown force constant. Show how we can predict the period of oscillation of this block spring system simple by measuring the extension of the spring produced by attaching the block is suspended from lower end of spring extension x is produced, when released it oscillates with time period.
Answer: Apply;
$$T=2\pi\sqrt{\frac{m}{k}}$$
$$T=2\pi\sqrt{\frac{m}{\sfrac{mg}{x}}}\Rightarrow2\pi\sqrt{\frac{x}{g}}$$
$$T\propto\sqrt x$$
This shows greater the extension ‘x’ greater will be time period T of system.
Q.20 A pendulum suspended from the ceiling of an elevator cab has period T when the cab is stationary. How its period is effected when the elevator moves
(a) – Upward with constant speed
(b) – Downward with constant speed
(c) – Downward with constant upward acceleration
(d) – Upward with constant upward acceleration
(e) – Upward with constant downward acceleration $\mathbit{a}$g.
(g) – In which case, if any does pendulum swing upside down.
Answer: The time period of simple pendulum is given by,
$$\mathbit{T}=\mathbf{2}\mathbit{\pi}\sqrt{\frac{\mathbit{l}}{\mathbit{g}}}$$
(a,b) – When can moves upward or downward with constant speed, in both the cases acceleration is zero (a = 0) due to which its time period remains the same.
( c ) – For downward motion, $T=2\pi\sqrt{\frac{l}{g-a}}$ g decreases to (g- a). Hence time period increases.
( d ) – For upward motion, $T=2\pi\sqrt{\frac{l}{g+a}}$ g increases to ( g+ a). Hence time period decreases.
( e ) – or upward motion with $a\ g\ ,\ \ T=2\pi\sqrt{\frac{l}{g-a}}$ which gives time period increases.
( g ) –When pendulum moves downward with $a>g$ the upside of the pendulum becomes down ( case f ).
Q.21 A singer, holding a note of right frequency shatter a glass, if the glassware is of high quality. This cannot be done, if glass ware quality is low. Explain, why, in terms of damping constant of glass.
Answer: High quality glass has low damping and it vibrate with large amplitude due to resonance with and finally breaks, due to increasing amplitude. Low quality glass has large damping and its amplitude of oscillation does not increase to the breaking limit.
